° above the horizontal. The tower is 50 meters tall. Calculate:

a) The time it takes for the ball to hit the ground.
b) The horizontal and vertical components of the ball's velocity just before it hits the ground.
c) The total distance the ball travels before hitting the ground.

a) To solve for the time it takes for the ball to hit the ground, we can use the formula:

h = ut + 0.5at^2

Where h is the initial height of the ball (50 meters), u is the initial velocity (10 m/s), a is the acceleration due to gravity (-9.81 m/s^2), and t is the time it takes for the ball to hit the ground (unknown).

Substituting the given values, we get:

50 = 10t + 0.5(-9.81)t^2

Rearranging and simplifying:

4.905t^2 + 10t - 50 = 0

Using the quadratic formula:

t = (-b ± sqrt(b^2 - 4ac))/2a

Where a = 4.905, b = 10, and c = -50.

Solving for t, we get:

t = 3.84 seconds (rounded to two decimal places)

Therefore, it takes the ball approximately 3.84 seconds to hit the ground.

b) To solve for the horizontal and vertical components of the ball's velocity just before it hits the ground, we can use the following equations:

v = u + at
v^2 = u^2 + 2as

Where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.

The vertical component of the ball's velocity just before it hits the ground can be found using the first equation, where a is the acceleration due to gravity:

v_vertical = u_vertical + at

Where u_vertical is the initial vertical component of the velocity, which is:

u_vertical = u*sin(30°) = 5 m/s

Substituting the given values, we get:

v_vertical = 5 + (-9.81)*3.84

v_vertical = -32.78 m/s (rounded to two decimal places)

The negative sign indicates that the ball is moving downwards.

The horizontal component of the ball's velocity remains constant throughout its flight, since there is no horizontal acceleration. Therefore, the horizontal component of the velocity just before it hits the ground is:

v_horizontal = u_horizontal = u*cos(30°) = 8.66 m/s

c) To solve for the total distance the ball travels before hitting the ground, we can use the following equation:

s = ut + 0.5at^2

We can solve for the horizontal and vertical components of the displacement separately, since they are perpendicular to each other.

The horizontal component of the displacement is:

s_horizontal = u_horizontal*t = 8.66*3.84 = 33.26 m

The vertical component of the displacement is:

s_vertical = u_vertical*t + 0.5at^2 = 5*3.84 + 0.5*(-9.81)*3.84^2 = -59.53 m

The negative sign indicates that the ball is moving downwards.

The total distance the ball travels before hitting the ground is the magnitude of its displacement,